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(1) is not sufficient because if both triangles have the same area, using the area formula, and given that both have the same height, D is the mid point of AC. However, knowing this you cannot tell if the triangle is isso.
INSUFFICIENT

(2) SUFFICIENT because using Pyt theorem, the hyp of each triangle must be the same.

(1) is not sufficient because if both triangles have the same area, using the area formula, and given that both have the same height, D is the mid point of AC. However, knowing this you cannot tell if the triangle is isso. INSUFFICIENT

(2) SUFFICIENT because using Pyt theorem, the hyp of each triangle must be the same.

For I, if the areas are the same, then D is the midpoint of AC right. So if AD= CD, then BD would have bisected AC. So isn't BD an angle bisector of angle ABC as well?

(1) is not sufficient because if both triangles have the same area, using the area formula, and given that both have the same height, D is the mid point of AC. However, knowing this you cannot tell if the triangle is isso. INSUFFICIENT

(2) SUFFICIENT because using Pyt theorem, the hyp of each triangle must be the same.

For I, if the areas are the same, then D is the midpoint of AC right. So if AD= CD, then BD would have bisected AC. So isn't BD an angle bisector of angle ABC as well?

Yes, according to stat 1, angle ABD = angle DBC. But that does not make the triangle isoceles. That's only one angle of triangle ABD = one angle of triangle DBC.

Stat 1:
Since areas are equal: 1/2 *AD * h = 1/2 DC*h or AD=DC. But we cannot prove similarity or another angle of both triangles equal using this. Insuff.

(1) is not sufficient because if both triangles have the same area, using the area formula, and given that both have the same height, D is the mid point of AC. However, knowing this you cannot tell if the triangle is isso. INSUFFICIENT

(2) SUFFICIENT because using Pyt theorem, the hyp of each triangle must be the same.

For I, if the areas are the same, then D is the midpoint of AC right. So if AD= CD, then BD would have bisected AC. So isn't BD an angle bisector of angle ABC as well?

Yes, according to stat 1, angle ABD = angle DBC. But that does not make the triangle isoceles. That's only one angle of triangle ABD = one angle of triangle DBC.

Stat 1: Since areas are equal: 1/2 *AD * h = 1/2 DC*h or AD=DC. But we cannot prove similarity or another angle of both triangles equal using this. Insuff.

Stat 2: angle ABD = angle DBCangle BDA = angle BDCSuff.

B

But doesn't the height or the altitude have to be perpendicular to the base? Thereby creating a right triangle - ADB = 90=CDB??

(1) is not sufficient because if both triangles have the same area, using the area formula, and given that both have the same height, D is the mid point of AC. However, knowing this you cannot tell if the triangle is isso. INSUFFICIENT

(2) SUFFICIENT because using Pyt theorem, the hyp of each triangle must be the same.

For I, if the areas are the same, then D is the midpoint of AC right. So if AD= CD, then BD would have bisected AC. So isn't BD an angle bisector of angle ABC as well?

Yes, according to stat 1, angle ABD = angle DBC. But that does not make the triangle isoceles. That's only one angle of triangle ABD = one angle of triangle DBC.

Stat 1: Since areas are equal: 1/2 *AD * h = 1/2 DC*h or AD=DC. But we cannot prove similarity or another angle of both triangles equal using this. Insuff.

Stat 2: angle ABD = angle DBCangle BDA = angle BDCSuff.

B

But doesn't the height or the altitude have to be perpendicular to the base? Thereby creating a right triangle - ADB = 90=CDB??

The height does have to be perpendicular to the base--but BD isn't necessarily the height. If ADB and CDB aren't right angles, then you would have to drop a perpendicular from B to an extension of the line AC for the height. (Note that the height of a triangle isn't necessarily contained inside the triangle.)

Another way of looking at it: given statement 1, it's possible that BD is perpendicular to AC, in which case ABC is isosceles. But it's also possible that ABC looks just like the drawing, with AD = CD, but ADB < CDB, so that AB < CB.